Which Loss Transformations Preserve Dominance?

Why the Scale of Loss Is Different From Its Shape

A loss function assigns a numerical cost to a decision. Changing the unit in which that cost is measured should not change which procedure has the smaller risk: measuring a loss in cents rather than dollars multiplies every loss by the same positive constant. A more substantial change, such as squaring each realized loss, can behave differently. Risk averages loss over the data distribution, and in general averaging a nonlinear transformation is not the same as transforming an average.

The distinction matters for admissibility. Dominance compares procedures through their risk functions, one parameter value at a time. To assess a change of loss, we must therefore ask what happens to those risk functions—not just whether the transformed loss is larger or smaller for each individual loss value.

Definition: In a statistical decision problem, let $X$ have distribution $P_\theta$ when the parameter is $\theta$, let $\delta$ be a decision rule, and let $L(\theta,a)$ be the loss from taking action $a$. The risk of $\delta$ is $$ R_L(\delta,\theta)=E_\theta[L(\theta,\delta(X))]. $$ For a function $g$, the risk under the transformed loss $g\circ L$ is $$ R_{g\circ L}(\delta,\theta)=E_\theta[g(L(\theta,\delta(X)))]. $$ The latter is generally not $g(R_L(\delta,\theta))$.

Assume the relevant losses and risks are measurable and finite, so the expectations below exist. A rule $\delta$ dominates a rule $\eta$ under $L$ if $R_L(\delta,\theta)\leq R_L(\eta,\theta)$ for every $\theta$, with strict inequality for at least one $\theta$. If the inequalities are weak at every parameter, including equality everywhere, the rules are weakly ordered in that direction. These comparisons are pointwise over the parameter space.

Positive Rescaling Preserves the Risk Order

For a fixed constant $c>0$, define the rescaled loss $L_c(\theta,a)=cL(\theta,a)$. Since expectation is linear, the transformed risk is exactly the original risk multiplied by $c$. This identity gives preservation of both the weak comparisons and the strict comparisons that define dominance.

Theorem: Suppose $c>0$ is a finite constant and the risks under $L$ are finite. For any two rules $\delta,\eta$ and every $\theta$, $$ R_{L_c}(\delta,\theta)=cR_L(\delta,\theta). $$ Consequently, $\delta$ dominates $\eta$ under $L$ if and only if $\delta$ dominates $\eta$ under $L_c$. The same equivalence holds for weak risk order.

Proof. By the definition of risk and $L_c=cL$,

\begin{aligned} R_{L_c}(\delta,\theta) &=E_\theta[L_c(\theta,\delta(X))]\\ &=E_\theta[cL(\theta,\delta(X))]\\ &=cE_\theta[L(\theta,\delta(X))]\\ &=cR_L(\delta,\theta). \end{aligned}

If $R_L(\delta,\theta)\leq R_L(\eta,\theta)$, multiplying by $c>0$ gives $cR_L(\delta,\theta)\leq cR_L(\eta,\theta)$. The identity just proved turns this into $R_{L_c}(\delta,\theta)\leq R_{L_c}(\eta,\theta)$. The argument applies at every $\theta$. If the inequality is strict at some $\theta_0$, multiplying by $c>0$ leaves it strict there. Conversely, division by $c>0$ recovers every original comparison from a transformed one. This proves both directions and the strict part of dominance. $\square$

Worked Example: Changing the Unit of Loss

Suppose two rules have finite risks at two parameter values:

Rule	$R_L(\cdot,\theta_1)$	$R_L(\cdot,\theta_2)$
$\delta$	1	3
$\eta$	2	4

At $\theta_1$, $1<2$, and at $\theta_2$, $3<4$, so $\delta$ dominates $\eta$. Now express every loss in a unit one hundred times as large, which means $L_c=100L$. The risks become $R_{L_c}(\delta,\cdot)=(100,300)$ and $R_{L_c}(\eta,\cdot)=(200,400)$. The inequalities remain strict at both parameter values. The numbers changed, but the risk order did not.

The sign and size restrictions on the multiplier are important. If $c=0$, all transformed risks are zero, so strict comparisons disappear. If $c<0$, multiplication reverses inequalities rather than preserving them. Thus the result is specifically about a common, positive multiplier.

Why Nonlinear Transformations Can Change Dominance

For a nonlinear function $g$, the transformed risk is an expectation of $g$ applied to realized losses. Its value depends on the distribution of those losses, not just on their original mean. Therefore one rule can have a smaller mean loss but a larger expected transformed loss. This can happen even when $g$ is strictly increasing: increasing $g$ preserves the ordering of two fixed loss values, but risk comparisons average over possibly different distributions of loss.

Worked Example: Squaring Reverses a Dominance Comparison

There is one parameter value, $\theta=0$, and the observation $X$ is Bernoulli with $P_0(X=0)=P_0(X=1)=1/2$. The action space is $\{0,2.1,4\}$, and the loss is $L(0,a)=a$. Let $\delta$ choose action 0 when $X=0$ and action 4 when $X=1$. Let $\eta$ always choose action 2.1.

Under the original loss, $\delta$'s risk is $$ R_L(\delta,0)=\tfrac12(0)+\tfrac12(4)=2, $$ while $\eta$'s risk is $R_L(\eta,0)=2.1$. There is only one parameter value, and $2<2.1$, so $\delta$ dominates $\eta$.

Now transform the loss by $g(t)=t^2$, which is strictly increasing for $t\geq0$. The transformed risk of $\delta$ is $$ R_{g\circ L}(\delta,0)=\tfrac12(0^2)+\tfrac12(4^2)=8, $$ whereas the transformed risk of $\eta$ is $$ R_{g\circ L}(\eta,0)=2.1^2=4.41. $$ Thus $\eta$ has the smaller transformed risk. The original dominance comparison is reversed. In particular, $g(R_L(\delta,0))=2^2=4$, which is not the transformed risk $8$.

Worked Example: A Concave Transformation Also Reverses the Order

Again take one parameter value, $\theta=0$, with $X$ Bernoulli and both outcomes having probability $1/2$. Let the action space be $\{0,2,5\}$, with $L(0,a)=a$. Rule $\delta$ always takes action 2. Rule $\eta$ takes action 0 when $X=0$ and action 5 when $X=1$.

The original risks are $$ R_L(\delta,0)=2,\qquad R_L(\eta,0)=\tfrac12(0)+\tfrac12(5)=2.5. $$ Hence $\delta$ dominates $\eta$ under $L$. Transform losses using the strictly increasing function $g(t)=\sqrt{t}$ on $t\geq0$. The transformed risks are $$ R_{g\circ L}(\delta,0)=\sqrt{2}\approx1.414,\qquad R_{g\circ L}(\eta,0)=\tfrac12\sqrt{0}+\tfrac12\sqrt{5}\approx1.118. $$ Now $\eta$ has the smaller risk. The transformation is increasing, but it changes how variation in realized losses contributes to the average. Together, the squaring and square-root examples show that the failure is not limited to transformations that decrease as loss increases.

What Can Be Preserved Beyond Scaling?

A positive affine transformation $g(t)=at+b$, with $a>0$ and the same constant $b$ applied to every action and parameter, also preserves risk order when the expectations exist. Indeed,

R_{g\circ L}(\delta,\theta)=aR_L(\delta,\theta)+b.

Adding the same $b$ to both rules' risks at a given parameter does not change their ordering, and multiplication by $a>0$ preserves it. This is a useful extension, but it depends on applying the same shift everywhere. A parameter- or action-dependent addition need not cancel in comparisons.

By contrast, monotonicity of $g$ alone is not enough. It guarantees that if one realized loss is smaller than another fixed realized loss, their transformed values remain ordered. It does not guarantee that the expectations of transformed losses are ordered whenever the expectations of the original losses are. The examples demonstrate the precise gap between a comparison of fixed numbers and a comparison of averages over random outcomes.

Common Mistake

A tempting but incorrect argument is: “$g$ is increasing and $R_L(\delta,\theta)\leq R_L(\eta,\theta)$, so $g(R_L(\delta,\theta))\leq g(R_L(\eta,\theta))$; therefore dominance is preserved under transformed loss.” The inequality in that argument is true, but it compares transformed original risks. The risk under transformed loss is $E_\theta[g(L(\theta,\delta(X)))]$, which need not equal $g(E_\theta[L(\theta,\delta(X))])$. Squaring in the first example gives $8$ for the transformed risk of $\delta$, not $4$. To evaluate a transformed loss, transform each realized loss first and then take its expectation.

Key takeaway: A common positive rescaling multiplies every risk by the same positive constant, so it preserves weak and strict dominance in both directions. A nonlinear transformation acts on the distribution of realized loss before averaging; even a strictly increasing transformation can therefore reverse a risk comparison.

Check Your Understanding

Answer each question by distinguishing transformed loss from transformed risk.

If $L_c=cL$ for $c>0$, what is $R_{L_c}(\delta,\theta)$, and why does strict inequality remain strict?
What happens to strict dominance if every loss is multiplied by zero? What happens if every loss is multiplied by a negative constant?
In the squaring example, calculate $g(R_L(\delta,0))$ and compare it with $R_{g\circ L}(\delta,0)$. Why are they different?
Why does strict increase of $g$ on the range of losses fail to guarantee preservation of dominance?
State a condition on an affine transformation $g(t)=at+b$ that ensures it preserves risk order, and explain the role of the common constant $b$.

Rule	\(R_L(\cdot,\theta_1)\)	\(R_L(\cdot,\theta_2)\)
\(\delta\)	1	3
\(\eta\)	2	4